Some Basic Concepts Of Chemistry Level-2 Exam

1. Which has the maximum number of molecules among the following-

64 g SO2

44 g CO2

48 g O3

8 g H2

The correct answer is 8 g H2. The number of molecules in a substance is determined by its molar mass. Since H2 has the lowest molar mass among the given options, it will have the maximum number of molecules for a given mass.

Explanation

The correct answer is 8 g H2. The number of molecules in a substance is determined by its molar mass. Since H2 has the lowest molar mass among the given options, it will have the maximum number of molecules for a given mass.

2. 8 g NaOH is dissolved in one litre of solution. Its molarity is-

0.8 M

0.4 M

0.2 M

0.1 M

The molarity of a solution is calculated by dividing the number of moles of solute by the volume of the solution in liters. In this case, there are 8 grams of NaOH dissolved in one liter of solution. To find the number of moles, we need to convert grams to moles using the molar mass of NaOH. The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol. Therefore, the number of moles of NaOH is 8 g / 39.00 g/mol = 0.205 moles. Dividing this by the volume of the solution (1 L), we get a molarity of 0.205 M, which is closest to 0.2 M.

Explanation

The molarity of a solution is calculated by dividing the number of moles of solute by the volume of the solution in liters. In this case, there are 8 grams of NaOH dissolved in one liter of solution. To find the number of moles, we need to convert grams to moles using the molar mass of NaOH. The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol. Therefore, the number of moles of NaOH is 8 g / 39.00 g/mol = 0.205 moles. Dividing this by the volume of the solution (1 L), we get a molarity of 0.205 M, which is closest to 0.2 M.

3. In the reaction N₂ + 3H₂ → 2NH₃ Ratio by volume of N₂, H₂ and NH₃ is 1:3:2. This illustrates law of-

Definite proportion

Multiple proportion

Reciprocal proportion

Gay Lussac

The given reaction shows that for every 1 volume of N2, 3 volumes of H2 are required to produce 2 volumes of NH3. This ratio by volume is consistent with the law of Gay Lussac, which states that the ratio of volumes of gases in a chemical reaction is always in small whole numbers. Therefore, the correct answer is Gay Lussac.

Explanation

The given reaction shows that for every 1 volume of N2, 3 volumes of H2 are required to produce 2 volumes of NH3. This ratio by volume is consistent with the law of Gay Lussac, which states that the ratio of volumes of gases in a chemical reaction is always in small whole numbers. Therefore, the correct answer is Gay Lussac.

4. The element chlorine consists of a mixture of 75.53 % ₁₇Cl³⁵ and 24.47 % ₁₇Cl³⁷ having masses of 34.97 amu and 36.95 amu respectively. The atomic weight of chlorine is

35.25

35.45

36.25

36.45

The atomic weight of an element is the average mass of all its isotopes, taking into account their relative abundance. In this case, chlorine has two isotopes, 17Cl35 and 17Cl37, with masses of 34.97 amu and 36.95 amu respectively. The percentages given indicate the relative abundance of each isotope. To calculate the atomic weight, we multiply the mass of each isotope by its abundance, and then sum the results. In this case, (0.7553 * 34.97 amu) + (0.2447 * 36.95 amu) equals 35.45 amu, which is the correct answer.

Explanation

The atomic weight of an element is the average mass of all its isotopes, taking into account their relative abundance. In this case, chlorine has two isotopes, 17Cl35 and 17Cl37, with masses of 34.97 amu and 36.95 amu respectively. The percentages given indicate the relative abundance of each isotope. To calculate the atomic weight, we multiply the mass of each isotope by its abundance, and then sum the results. In this case, (0.7553 * 34.97 amu) + (0.2447 * 36.95 amu) equals 35.45 amu, which is the correct answer.

5. A compound is found to contain 80% carbon and 20% hydrogen. Which of the following is that compound?

C6H6

C2H5OH

C2H6

C2H4

The compound that contains 80% carbon and 20% hydrogen is C2H6, which is ethane. Ethane is a hydrocarbon that consists of two carbon atoms and six hydrogen atoms. This compound fits the given percentages of carbon and hydrogen, making it the correct answer.

Explanation

The compound that contains 80% carbon and 20% hydrogen is C2H6, which is ethane. Ethane is a hydrocarbon that consists of two carbon atoms and six hydrogen atoms. This compound fits the given percentages of carbon and hydrogen, making it the correct answer.

6. Which of the following pairs of compounds illustrates the law of multiple proportions?

H2O , Na2O

MgO , Na2O

Na2O , BaO

SnCl2 , SnCl4

The law of multiple proportions states that when two elements combine to form different compounds, the ratio of the masses of one element that combine with a fixed mass of the other element will always be a ratio of small whole numbers. In the given options, the pair SnCl2 and SnCl4 illustrates the law of multiple proportions because they both contain the same elements, tin (Sn) and chlorine (Cl), but in different ratios. The ratio of Sn to Cl in SnCl2 is 1:2, while in SnCl4 it is 1:4, which is a ratio of small whole numbers.

Explanation

The law of multiple proportions states that when two elements combine to form different compounds, the ratio of the masses of one element that combine with a fixed mass of the other element will always be a ratio of small whole numbers. In the given options, the pair SnCl2 and SnCl4 illustrates the law of multiple proportions because they both contain the same elements, tin (Sn) and chlorine (Cl), but in different ratios. The ratio of Sn to Cl in SnCl2 is 1:2, while in SnCl4 it is 1:4, which is a ratio of small whole numbers.

7. A gaseous hydrocarbon gives upon combustion, 0.72 g H₂O and 3.08 g CO₂. The empirical formula of the hydrocarbon is-

C2H4

C3H4

C6H5

C7H8

The given information states that upon combustion, the hydrocarbon produces 0.72 g of water and 3.08 g of carbon dioxide. To find the empirical formula, we need to determine the ratio of carbon to hydrogen in the compound. By calculating the molar masses of water and carbon dioxide, we find that the ratio of carbon to hydrogen is approximately 1:1. Therefore, the empirical formula of the hydrocarbon is C1H1, which can be simplified to CH. However, none of the given options match this empirical formula, so the correct answer is not available.

Explanation

The given information states that upon combustion, the hydrocarbon produces 0.72 g of water and 3.08 g of carbon dioxide. To find the empirical formula, we need to determine the ratio of carbon to hydrogen in the compound. By calculating the molar masses of water and carbon dioxide, we find that the ratio of carbon to hydrogen is approximately 1:1. Therefore, the empirical formula of the hydrocarbon is C1H1, which can be simplified to CH. However, none of the given options match this empirical formula, so the correct answer is not available.

8. Molecular weight of tribasic acid is W. Its equivalent weight will be-

W/2

W/3

W

3W

The equivalent weight of a substance is defined as the molecular weight divided by the number of acidic or basic equivalents present in the molecule. In the case of tribasic acid, it has three acidic equivalents. Therefore, the equivalent weight will be the molecular weight divided by three, which is W/3.

Explanation

The equivalent weight of a substance is defined as the molecular weight divided by the number of acidic or basic equivalents present in the molecule. In the case of tribasic acid, it has three acidic equivalents. Therefore, the equivalent weight will be the molecular weight divided by three, which is W/3.

9. Cl₂ reacts completely with 0.4 gram H₂ to yield HCl. The amount of HCl formed is-

7.3 grams

10.7 grams

14.4 grams

16.7 grams

When Cl2 reacts with H2, it forms HCl. The balanced chemical equation for this reaction is:

Cl2 + H2 -> 2HCl

From the equation, we can see that 1 mole of Cl2 reacts with 1 mole of H2 to produce 2 moles of HCl. To find the amount of HCl formed, we need to calculate the number of moles of H2.

Given that the mass of H2 is 0.4 grams, we can use the molar mass of H2 (2 grams/mole) to calculate the number of moles:

moles of H2 = mass of H2 / molar mass of H2 = 0.4 g / 2 g/mol = 0.2 mol

Since the reaction is 1:1 between Cl2 and H2, the number of moles of HCl formed will also be 0.2 moles.

To convert moles of HCl to grams, we can use the molar mass of HCl (36.5 grams/mole):

mass of HCl = moles of HCl * molar mass of HCl = 0.2 mol * 36.5 g/mol = 7.3 grams

Therefore, the amount of HCl formed is 7.3 grams.

Explanation

When Cl2 reacts with H2, it forms HCl. The balanced chemical equation for this reaction is:

Cl2 + H2 -> 2HCl

From the equation, we can see that 1 mole of Cl2 reacts with 1 mole of H2 to produce 2 moles of HCl. To find the amount of HCl formed, we need to calculate the number of moles of H2.

Given that the mass of H2 is 0.4 grams, we can use the molar mass of H2 (2 grams/mole) to calculate the number of moles:

moles of H2 = mass of H2 / molar mass of H2 = 0.4 g / 2 g/mol = 0.2 mol

Since the reaction is 1:1 between Cl2 and H2, the number of moles of HCl formed will also be 0.2 moles.

To convert moles of HCl to grams, we can use the molar mass of HCl (36.5 grams/mole):

mass of HCl = moles of HCl * molar mass of HCl = 0.2 mol * 36.5 g/mol = 7.3 grams

Therefore, the amount of HCl formed is 7.3 grams.

10. The number of atoms in 0.1 mol of a triatomic gas is ( N_A = 6.02 x 10²³mol^-1)

1.800 x 1022

6.026 x 1022

1.806 x 1023

3.600 x 1023

The number of atoms in 0.1 mol of a triatomic gas can be calculated by multiplying the Avogadro's number (6.02 x 10^23) by the number of moles. Therefore, the correct answer is 1.806 x 10^23.

Explanation

The number of atoms in 0.1 mol of a triatomic gas can be calculated by multiplying the Avogadro's number (6.02 x 10^23) by the number of moles. Therefore, the correct answer is 1.806 x 10^23.

11. A reaction is given as follows: SnO₂ + C → Sn + CO Its balanced form will involve

2C and 2CO

C and 3CO

C and 2CO

2C and CO

In order to balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this reaction, we have one Sn atom on the left side and one Sn atom on the right side, so that is already balanced. However, we have two O atoms on the left side and only one O atom on the right side, so we need to balance that by having two CO molecules on the right side. This means that we have two C atoms and two CO molecules on the right side, which matches the answer choice of 2C and 2CO.

Explanation

In order to balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this reaction, we have one Sn atom on the left side and one Sn atom on the right side, so that is already balanced. However, we have two O atoms on the left side and only one O atom on the right side, so we need to balance that by having two CO molecules on the right side. This means that we have two C atoms and two CO molecules on the right side, which matches the answer choice of 2C and 2CO.

12. How much volume of 3.0 M H₂SO₄ is required for preparation of 1.0 litre of 1.0 M solution?

300 mL

320 mL

333.33 mL

350 mL

To determine the volume of 3.0 M H2SO4 required for a 1.0 M solution, we can use the formula M1V1 = M2V2. We know that the initial concentration (M1) is 3.0 M, the initial volume (V1) is unknown, the final concentration (M2) is 1.0 M, and the final volume (V2) is 1.0 L. Plugging these values into the formula, we can solve for V1. Rearranging the formula, V1 = (M2V2) / M1 = (1.0 M * 1.0 L) / 3.0 M = 0.333 L = 333.33 mL. Therefore, 333.33 mL of 3.0 M H2SO4 is required for the preparation of 1.0 L of 1.0 M solution.

Explanation

To determine the volume of 3.0 M H2SO4 required for a 1.0 M solution, we can use the formula M1V1 = M2V2. We know that the initial concentration (M1) is 3.0 M, the initial volume (V1) is unknown, the final concentration (M2) is 1.0 M, and the final volume (V2) is 1.0 L. Plugging these values into the formula, we can solve for V1. Rearranging the formula, V1 = (M2V2) / M1 = (1.0 M * 1.0 L) / 3.0 M = 0.333 L = 333.33 mL. Therefore, 333.33 mL of 3.0 M H2SO4 is required for the preparation of 1.0 L of 1.0 M solution.

13. Given P = 0.0030 m , Q = 2.40 m , R = 3000 m . The significant figures for P,Q and R are respectively

2, 2, 1

2, 3, 4

4, 2, 1

4, 2, 4

The significant figures for P, Q, and R are respectively 2, 3, and 4. This is because significant figures are the digits in a number that carry meaning in terms of precision. In P, there are two significant figures (0 and 3) because the zeros before the 3 are not significant. In Q, there are three significant figures (2, 4, and 0) because all the digits are non-zero. In R, there are four significant figures (3, 0, 0, and 0) because all the zeros between the non-zero digits are significant.

Explanation

The significant figures for P, Q, and R are respectively 2, 3, and 4. This is because significant figures are the digits in a number that carry meaning in terms of precision. In P, there are two significant figures (0 and 3) because the zeros before the 3 are not significant. In Q, there are three significant figures (2, 4, and 0) because all the digits are non-zero. In R, there are four significant figures (3, 0, 0, and 0) because all the zeros between the non-zero digits are significant.

14. The molarity of solution obtained by mixing 200 mL 0.1 M HCl with 300 mL 0.2 M HCl solution will be -

0.08 M

0.20 M

0.06 M

0.16 M

When two solutions are mixed together, the final molarity can be calculated using the formula:

M1V1 + M2V2 = M3V3

Where M1 and M2 are the molarities of the two solutions, V1 and V2 are the volumes of the two solutions, and M3 is the final molarity.

In this case, the first solution has a molarity of 0.1 M and a volume of 200 mL, while the second solution has a molarity of 0.2 M and a volume of 300 mL.

Using the formula, we can calculate:

(0.1 M)(200 mL) + (0.2 M)(300 mL) = M3(200 mL + 300 mL)

20 mL + 60 mL = M3(500 mL)

80 mL = M3(500 mL)

M3 = 80 mL / 500 mL

M3 = 0.16 M

Therefore, the molarity of the solution obtained by mixing the two solutions will be 0.16 M.

Explanation

When two solutions are mixed together, the final molarity can be calculated using the formula:

M1V1 + M2V2 = M3V3

Where M1 and M2 are the molarities of the two solutions, V1 and V2 are the volumes of the two solutions, and M3 is the final molarity.

In this case, the first solution has a molarity of 0.1 M and a volume of 200 mL, while the second solution has a molarity of 0.2 M and a volume of 300 mL.

Using the formula, we can calculate:

(0.1 M)(200 mL) + (0.2 M)(300 mL) = M3(200 mL + 300 mL)

20 mL + 60 mL = M3(500 mL)

80 mL = M3(500 mL)

M3 = 80 mL / 500 mL

M3 = 0.16 M

Therefore, the molarity of the solution obtained by mixing the two solutions will be 0.16 M.

15. How many moles of magnesium phosphate Mg₃(PO₄)₂will contain 0.25 mole of oxygen atoms?

2.5 x 10-2

0.02

3.125 x 10-2

1.25 x 10-2

The molar ratio between oxygen atoms and magnesium phosphate is 10:2. This means that for every 2 moles of magnesium phosphate, there are 10 moles of oxygen atoms. Therefore, if there are 0.25 moles of oxygen atoms, there must be (0.25/10) x 2 = 0.05 moles of magnesium phosphate. In scientific notation, this is equivalent to 5 x 10-2 moles.

Explanation

The molar ratio between oxygen atoms and magnesium phosphate is 10:2. This means that for every 2 moles of magnesium phosphate, there are 10 moles of oxygen atoms. Therefore, if there are 0.25 moles of oxygen atoms, there must be (0.25/10) x 2 = 0.05 moles of magnesium phosphate. In scientific notation, this is equivalent to 5 x 10-2 moles.

16. Vapour density of ammonia is 8.5, 85 grams of NH₃ at STP will occupy-

22.4 litre

112 litre

224 litre

1120 litre

The vapour density of a substance is the ratio of its molar mass to the molar mass of hydrogen. In the case of ammonia (NH3), its molar mass is approximately 17 grams/mol. Since the vapour density is given as 8.5, it means that the molar mass of ammonia is 8.5 times that of hydrogen. Therefore, the molar mass of ammonia is 8.5 * 2 grams/mol (molar mass of hydrogen) = 17 grams/mol. At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. Hence, 85 grams of ammonia (which is approximately 5 moles) will occupy 5 * 22.4 liters = 112 liters.

Explanation

The vapour density of a substance is the ratio of its molar mass to the molar mass of hydrogen. In the case of ammonia (NH3), its molar mass is approximately 17 grams/mol. Since the vapour density is given as 8.5, it means that the molar mass of ammonia is 8.5 times that of hydrogen. Therefore, the molar mass of ammonia is 8.5 * 2 grams/mol (molar mass of hydrogen) = 17 grams/mol. At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. Hence, 85 grams of ammonia (which is approximately 5 moles) will occupy 5 * 22.4 liters = 112 liters.

17. A person adds 1.71 gram of sugar (C₁₂H₂₂O₁₁) in order to sweeten his tea. The number of carbon atoms added are- (Molecular mass of sugar = 342 g/mol)

3.6 x 1022

7.2 x 1021

0.05

6.6 x 1022

not-available-via-ai

Explanation

not-available-via-ai

18. According to the following reaction- 2Al + 1.5O₂ → Al₂O₃ 9 gms of Al will react completely with-

6 gms of O2

8 gms of O2

9 gms of O2

4 gms O2

In the given reaction, the stoichiometric ratio between Al and O2 is 2:1. This means that for every 2 moles of Al, 1 mole of O2 is required. The molar mass of Al is 27 g/mol, so 9 g of Al is equal to 9/27 = 1/3 moles. According to the stoichiometry, 1/3 moles of Al will react with 1/3 * 1/2 = 1/6 moles of O2. The molar mass of O2 is 32 g/mol, so 1/6 moles of O2 is equal to 1/6 * 32 = 8 g. Therefore, 9 g of Al will react completely with 8 g of O2.

Explanation

In the given reaction, the stoichiometric ratio between Al and O2 is 2:1. This means that for every 2 moles of Al, 1 mole of O2 is required. The molar mass of Al is 27 g/mol, so 9 g of Al is equal to 9/27 = 1/3 moles. According to the stoichiometry, 1/3 moles of Al will react with 1/3 * 1/2 = 1/6 moles of O2. The molar mass of O2 is 32 g/mol, so 1/6 moles of O2 is equal to 1/6 * 32 = 8 g. Therefore, 9 g of Al will react completely with 8 g of O2.

19. The molarity of 0.2 N Na₂CO₃ solution will be-

0.05 M

0.2 M

0.1 M

0.4 M

The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the solution is 0.2 N Na2CO3, which means it is a 0.2 normal solution. A normal solution is one that contains 1 gram equivalent weight of solute per liter of solution. To find the molarity, we need to convert the normality to molarity by dividing by the equivalent weight of Na2CO3, which is 106 g/mol. Therefore, 0.2 N Na2CO3 is equivalent to 0.1 M Na2CO3.

Explanation

The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the solution is 0.2 N Na2CO3, which means it is a 0.2 normal solution. A normal solution is one that contains 1 gram equivalent weight of solute per liter of solution. To find the molarity, we need to convert the normality to molarity by dividing by the equivalent weight of Na2CO3, which is 106 g/mol. Therefore, 0.2 N Na2CO3 is equivalent to 0.1 M Na2CO3.

20. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction is-

1 mol

2 mol

3 mol

4 mol

In this reaction, hydrogen and oxygen combine to form water according to the balanced equation: 2H₂ + O₂ → 2H₂O. From the given amounts of hydrogen (10g) and oxygen (64g), we can calculate the number of moles of each element. The molar mass of hydrogen is 2 g/mol, so 10g of hydrogen is equivalent to 5 moles. The molar mass of oxygen is 32 g/mol, so 64g of oxygen is equivalent to 2 moles. According to the balanced equation, for every 2 moles of hydrogen, 2 moles of water are produced. Therefore, for 5 moles of hydrogen, 5 moles of water are produced. Hence, the amount of water produced in this reaction is 4 mol.

Explanation

In this reaction, hydrogen and oxygen combine to form water according to the balanced equation: 2H₂ + O₂ → 2H₂O. From the given amounts of hydrogen (10g) and oxygen (64g), we can calculate the number of moles of each element. The molar mass of hydrogen is 2 g/mol, so 10g of hydrogen is equivalent to 5 moles. The molar mass of oxygen is 32 g/mol, so 64g of oxygen is equivalent to 2 moles. According to the balanced equation, for every 2 moles of hydrogen, 2 moles of water are produced. Therefore, for 5 moles of hydrogen, 5 moles of water are produced. Hence, the amount of water produced in this reaction is 4 mol.

21. When 100 g of ethylene polymerizes to polythene according to the equation- n CH₂=CH₂ → -(-CH₂ - CH₂ -)_n The weight of polythene produced will be-

N/2 grams

100 grams

100/n grams

100n grams

When 100 g of ethylene polymerizes to polythene, the equation indicates that each ethylene molecule (-CH2-CH2-) combines to form a polymer chain (-(-CH2-CH2-)n). The "n" in the equation represents the number of ethylene molecules that have polymerized. Since the equation does not provide any information about the value of "n," we cannot determine the exact weight of polythene produced. Therefore, the weight of polythene produced cannot be determined based on the given information.

Explanation

When 100 g of ethylene polymerizes to polythene, the equation indicates that each ethylene molecule (-CH2-CH2-) combines to form a polymer chain (-(-CH2-CH2-)n). The "n" in the equation represents the number of ethylene molecules that have polymerized. Since the equation does not provide any information about the value of "n," we cannot determine the exact weight of polythene produced. Therefore, the weight of polythene produced cannot be determined based on the given information.

22. The mole fraction of oxygen in a mixture of 7 g nitrogen and 8 g oxygen is -

8/15

0.5

0.25

1.0

The mole fraction of a component in a mixture is calculated by dividing the moles of that component by the total moles of all components in the mixture. In this case, we can calculate the moles of nitrogen and oxygen using their respective molar masses. The molar mass of nitrogen is 28 g/mol and the molar mass of oxygen is 32 g/mol. Therefore, the moles of nitrogen are 7 g / 28 g/mol = 0.25 mol, and the moles of oxygen are 8 g / 32 g/mol = 0.25 mol. The total moles of both components in the mixture is 0.25 mol + 0.25 mol = 0.5 mol. Therefore, the mole fraction of oxygen is 0.25 mol / 0.5 mol = 0.5.

Explanation

The mole fraction of a component in a mixture is calculated by dividing the moles of that component by the total moles of all components in the mixture. In this case, we can calculate the moles of nitrogen and oxygen using their respective molar masses. The molar mass of nitrogen is 28 g/mol and the molar mass of oxygen is 32 g/mol. Therefore, the moles of nitrogen are 7 g / 28 g/mol = 0.25 mol, and the moles of oxygen are 8 g / 32 g/mol = 0.25 mol. The total moles of both components in the mixture is 0.25 mol + 0.25 mol = 0.5 mol. Therefore, the mole fraction of oxygen is 0.25 mol / 0.5 mol = 0.5.

23. The total number of ions present in 1 mL of 0.1 M barium nitrate Ba(NO₃)₂solution is-

6.02 x 1018

7.2 x 1021

3.0 x 6.02 x 1019

3.0 x 6.02 x 1018

The formula for barium nitrate is Ba(NO3)2. This means that for every one molecule of barium nitrate, there are two nitrate ions (NO3-) present. Since the concentration of the solution is 0.1 M, this means that there are 0.1 moles of barium nitrate in 1 L of solution. To convert this to 1 mL, we divide by 1000, giving us 0.0001 moles of barium nitrate. Since there are two nitrate ions for every one molecule of barium nitrate, we multiply the number of moles of barium nitrate by 2 to get the total number of nitrate ions. Therefore, the total number of ions present in 1 mL of 0.1 M barium nitrate solution is 3.0 x 6.02 x 10^19.

Explanation

The formula for barium nitrate is Ba(NO3)2. This means that for every one molecule of barium nitrate, there are two nitrate ions (NO3-) present. Since the concentration of the solution is 0.1 M, this means that there are 0.1 moles of barium nitrate in 1 L of solution. To convert this to 1 mL, we divide by 1000, giving us 0.0001 moles of barium nitrate. Since there are two nitrate ions for every one molecule of barium nitrate, we multiply the number of moles of barium nitrate by 2 to get the total number of nitrate ions. Therefore, the total number of ions present in 1 mL of 0.1 M barium nitrate solution is 3.0 x 6.02 x 10^19.

24. If 18 g of glucose is present in 1000 g solvent, the solution is said to be-

1 molar

0.1 molar

0.5 molar

0.1 molal

The given question is asking about the concentration of the solution. In this case, the concentration is expressed in terms of molality (molal). Molality is defined as the number of moles of solute per kilogram of solvent. The solution is said to be 0.1 molal because there are 0.1 moles of glucose present in 1 kilogram of solvent.

Explanation

The given question is asking about the concentration of the solution. In this case, the concentration is expressed in terms of molality (molal). Molality is defined as the number of moles of solute per kilogram of solvent. The solution is said to be 0.1 molal because there are 0.1 moles of glucose present in 1 kilogram of solvent.

25. The maximum number of molecules is present in-

5 L of N2 gas at STP

0.5 g of H2 gas

10 g of O2 gas

15 L of H2 gas at STP

The maximum number of molecules is present in 15 L of H2 gas at STP because Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore, the larger the volume of gas, the greater the number of molecules it can contain. Additionally, H2 gas has a molar mass of 2 g/mol, so 15 L of H2 gas at STP would have a greater number of molecules compared to the other options.

Explanation

The maximum number of molecules is present in 15 L of H2 gas at STP because Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore, the larger the volume of gas, the greater the number of molecules it can contain. Additionally, H2 gas has a molar mass of 2 g/mol, so 15 L of H2 gas at STP would have a greater number of molecules compared to the other options.

26. Number of atoms in 560 g Fe (atomic mass of Fe = 56 g/mol) is

Twice that of 70 g N

Half that of 20 g H

Both (a) and (b)

None of these

The number of atoms in a substance can be calculated using Avogadro's number and the molar mass of the substance. In this case, the molar mass of Fe is 56 g/mol and the molar mass of N is 14 g/mol.

For option (a), we have 560 g of Fe. To find the number of moles, we divide the mass by the molar mass: 560 g / 56 g/mol = 10 mol. Since 1 mole of Fe contains 6.02 x 10^23 atoms (Avogadro's number), we multiply the number of moles by Avogadro's number: 10 mol x 6.02 x 10^23 atoms/mol = 6.02 x 10^24 atoms.

For option (b), we have 70 g of N. Using the same calculations as above, we find that there are 3.51 x 10^24 atoms in 70 g of N.

Therefore, the number of atoms in 560 g Fe is twice that of 70 g N, and half that of 20 g H. Hence, the correct answer is both (a) and (b).

Explanation

The number of atoms in a substance can be calculated using Avogadro's number and the molar mass of the substance. In this case, the molar mass of Fe is 56 g/mol and the molar mass of N is 14 g/mol.

For option (a), we have 560 g of Fe. To find the number of moles, we divide the mass by the molar mass: 560 g / 56 g/mol = 10 mol. Since 1 mole of Fe contains 6.02 x 10^23 atoms (Avogadro's number), we multiply the number of moles by Avogadro's number: 10 mol x 6.02 x 10^23 atoms/mol = 6.02 x 10^24 atoms.

For option (b), we have 70 g of N. Using the same calculations as above, we find that there are 3.51 x 10^24 atoms in 70 g of N.

Therefore, the number of atoms in 560 g Fe is twice that of 70 g N, and half that of 20 g H. Hence, the correct answer is both (a) and (b).

27. Volume occupied by one molecule of water (density = 1 g/cm³) is-

3.0 x 10-23 cm3

5.5 x 10-23 cm3

9.0 x 10-23 cm3

6.023 x 10-23 cm3

The volume occupied by one molecule of water can be calculated using the formula: volume = mass/density. Since the density of water is given as 1 g/cm3, and the mass of one molecule of water is negligible, the volume occupied by one molecule of water would be very small. Among the given options, the volume closest to this value is 3.0 x 10-23 cm3.

Explanation

The volume occupied by one molecule of water can be calculated using the formula: volume = mass/density. Since the density of water is given as 1 g/cm3, and the mass of one molecule of water is negligible, the volume occupied by one molecule of water would be very small. Among the given options, the volume closest to this value is 3.0 x 10-23 cm3.

28. If 3.01 x 10²⁰ molecules are removed from 98 mg of H₂SO_4, then the number of moles of H₂SO₄ left are-

0.1 x 10-3

0.5 x 10-3

1.66 x 10-3

9.95 x 10-2

The given question involves calculating the number of moles of H2SO4 left after a certain number of molecules are removed. To solve this, we need to convert the given mass of H2SO4 (98 mg) into moles using its molar mass. Then, we can use the Avogadro's number to convert the number of molecules removed (3.01 x 10^20) into moles. Finally, subtracting the moles of molecules removed from the initial moles of H2SO4 will give us the moles of H2SO4 left. The correct answer of 0.5 x 10^-3 represents the moles of H2SO4 left after the molecules are removed.

Explanation

The given question involves calculating the number of moles of H2SO4 left after a certain number of molecules are removed. To solve this, we need to convert the given mass of H2SO4 (98 mg) into moles using its molar mass. Then, we can use the Avogadro's number to convert the number of molecules removed (3.01 x 10^20) into moles. Finally, subtracting the moles of molecules removed from the initial moles of H2SO4 will give us the moles of H2SO4 left. The correct answer of 0.5 x 10^-3 represents the moles of H2SO4 left after the molecules are removed.

29. For the reaction given below- A + 2B → C 5 moles of A and 8 moles of B will produce -

5 moles of C

4 moles of C

8 mole of C

13 mole of C

In the balanced chemical equation A + 2B → C, it is stated that for every 1 mole of A, 2 moles of B are required to produce 1 mole of C. Therefore, if there are 5 moles of A and 8 moles of B, there is an excess of B. Since B is the limiting reactant, it will determine the amount of C produced. Since 2 moles of B produce 1 mole of C, 8 moles of B will produce 4 moles of C. Therefore, the correct answer is 4 moles of C.

Explanation

In the balanced chemical equation A + 2B → C, it is stated that for every 1 mole of A, 2 moles of B are required to produce 1 mole of C. Therefore, if there are 5 moles of A and 8 moles of B, there is an excess of B. Since B is the limiting reactant, it will determine the amount of C produced. Since 2 moles of B produce 1 mole of C, 8 moles of B will produce 4 moles of C. Therefore, the correct answer is 4 moles of C.

30. In a gaseous reaction of the type- aA + bB → cC + dD Identify the wrong statement.

A litre of A combines with b litre of B to give C and D

A mole of A combines with b mole of B to give C and D

A gram of A combines with b gram of B to give C and D

A molecules of A combines with b molecules of B to give C and D

The given statement "a gram of A combines with b gram of B to give C and D" is incorrect. In a chemical reaction, the stoichiometric coefficients represent the ratio of moles, not grams. Therefore, the correct statement should be "a mole of A combines with b mole of B to give C and D."

Explanation

The given statement "a gram of A combines with b gram of B to give C and D" is incorrect. In a chemical reaction, the stoichiometric coefficients represent the ratio of moles, not grams. Therefore, the correct statement should be "a mole of A combines with b mole of B to give C and D."